[bonanova]

Here are two related challenges; one simple, one more difficult

Challenge 1:

A magician produces a deck of four cards, 2 red and 2 black, and deals them face down [the backs of the cards are indistinguishable.]

He bets that you cannot point to two cards of the same color, and offers even odds.

1. A friend urges you to take the bet:
   "There are only three cases: 2 red, 2 black and one each.
   You win 2 out of 3 times."
   .
2. Another friend says you will neither win nor lose:
   "There are four cases: red-red, red-black, black-red, and black black.
   You win 2 out of 4 times, and that just matches the odds."

Quickly decide which friend is right.

Challenge 2:

A deck with equal numbers of red and black cards gives a probability p that two selected cards will have the same color.

If a card is removed the color symmetry is broken, and one color is more likely to be represented in the two chosen cards.

Calculate the increase in p resulting from the removal of the card.

Enjoy, and please use spoilers

[Guest]

challenge 1

You win 1 out of three. Point to a card. Given that that is your choice, even though you don't know what it is, one of the remaining three cards will match.

[Guest]

challenge 1

You win 1 out of three. Point to a card. Given that that is your choice, even though you don't know what it is, one of the remaining three cards will match.

times two colors is two out of three.

[Guest]

fren 1.1 each means that red-black,black-red, which is also the same.

[Guest]

There are 6 ways two cards can be chosen from four, only 2 of which will give 2 cards of the same color, giving a probability of one out of three. Neither friend is right.

[Guest]

neither is right, you win in 1/3 of cases

according to my calculation p doesn't change. p=(t-1)/(2t-1) while t is the number of cards those are in unremoved color.

[Pickett]

neither is right, you win in 1/3 of cases

according to my calculation p doesn't change. p=(t-1)/(2t-1) while t is the number of cards those are in unremoved color.

So, I agree with these answers...but I decided to put the math and proof down for this:

#1 is just a specific implementation of the original case (not removing a card)...so let's say that n is the number of red and black cards in the deck...which means that the total number of cards in the deck is 2n...

After you pick a card, you have a ^(n-1)/_(2n-1) chance of picking the same color card (so p = ^(n-1)/_(2n-1))

Now, if we take one card away at the beginning (let's say red for the sake of argument...but it could be either) we are left with 2n-1 cards in the deck: n-1 red, n black.

Now let's say you pick a black card first:

you now have a 50% chance of matching the colors...^(n-1)/_(2n-2)

if you choose red card first:

you have a ^(n-2)/_(2n-2) probability...

So, your total probability for this scenario is:

p = ^(n-1)/_(2n-1)*^(n-2)/_(2n-2) + ⁿ/_(2n-1)*^(n-1)/_(2n-2)

Now it's just simple algebra:

^{(n^2-3n+2)}/_(4n^2-6n+2) + ^{(n^2-n)}/_(4n^2-6n+2)

^{(2n^2-4n+2)}/_(4n^2-6n+2)

^(2n-2)(n-1)/_(4n-2)(n-1)

^(2n-2)/_(4n-2)

^(n-1)/_(2n-1)

Which is the exact same probability that we had in case #1...and it matches what nobody had as his answer. Nice problem.

[Guest]

..maybe a little simpler,

The number of ways of picking 2 cards are: (2n-1)(2n-2)/2! = (2n-1)(n-1)

The number of ways of picking 2 distinct cards are: n(n-1)

So the number of ways of picking 2 cards of same color = (2n-1)(n-1) - n(n-1) = (n-1)(n-1)

So the probability is, (n-1)(n-1)/(2n-1)(n-1) = (n-1)/(2n-1)

Very curious result, would have expected the probability to have gone up. Obviously, if we keep discarding cards of the same color from the deck, the probability must increase (and become 1 ultimately).

[Guest]

both friends are wrong as there are six cases. two with same colours and four with different colours

[Guest]

There will be no change in p

[Guest]

assign the letters R,r,B,b to each of the cards. You will quickly see that the following combinations are possible: Rr, RB, Rb, rB, rb, Bb. Of these, one combination has 2 red cards, and one combination has 2 black cards. 4 combinations have 1 red and 1 black. The probability is 2/6, or 1/3 that they will be the same color. Do not take the bet.

When there are R red cards and B black cards, the probability of drawing 2 red cards is R*(R-1)/((R+B)*(R+B-1)). Likewise, the probability of drawing 2 black cards is B*(B-1)/((R+B)*(R+B-1)). If R and B are equal, then the probability of drawing 2 cards the same color is 2*R*(R-1)/(2*R)*(2*R-1), or (R-1)/(2R-1). If you remove a card from the deck (let's say it's red, leaving you with r = R-1 red cards), then you have (r*(r-1)+R*(R-1))/((r+R)*(r+R-1)), or ((R-1)*(R-2)+R*(R-1))/((2R-1)*(2R-2)), which simplifies to (R-1)/(2R-1), having exactly the same probability.

From spoiler 1, you can see that removing the R card, you now have the following possibilities: rB, rb, Bb, or a 1/3 chance.

[bonanova]

Nice job on Challenge 2.

Very counter-intuitive result.